Question

$E_{Cu}^{\circ }=0.34V,E_{Zn}^{\circ }=-0.76V.A$ Daniell cell contains $0.1MZnS{O}_4$ solution and $0.01M$ $CuS{O}_4$ solution at its electrodes. E.M.F. of the cell is

• A. 1.10 V
• B. 1.04 V
• C. 1.16 V
• D. 1.07 V

Answer 1

Answer: (D)

$Zn|Z{n}^{2+}||C{u}^{2+}|Cu$
$Zn+C{u}^{2+}\to Z{n}^{2+}+Cu$
$E_{cell}^0=E_{C{u}^{2+}/Cu}^0-E_{Z{n}^{2+}/Zn}^0$
$E_{cell}=E_{cell}^0-\frac{0.0591}{2}log\frac{[Z{n}^{2+}]}{[C{u}^{2+}]}$
$=1.10V-\frac{0.0591V}{2}log\frac{0.1}{0.01}$
$=1.10V-0.05955V$
$=1.07045V$